There are five problems. Solve four of the five problems. Each problem is worth 25 points. A sheet of convenient formulae is provided.

Transcription

1 Preliminary Examination (Solutions): Partial Differential Equations, 1 AM - 1 PM, Jan. 18, 16, oom Discovery Learning Center (DLC) Bechtel Collaboratory. Student ID: There are five problems. Solve four of the five problems. Each problem is worth 5 points. A sheet of convenient formulae is provided. # possible score Total 1 1. (Solution methods) Let Ω = (, π) (, T ), T >. Consider the initial, boundary value problem u t = u xx, < x < π, t > u x (π, t) =, u(, t) =, t >, u(x, ) = φ(x), < x < π. (a) Find a formal solution u(x, t) that solves the above initial value problem. (b) Find sufficient conditions on φ such that the formal solution u is classical, i.e., it is in C 1( Ω), functions that are twice continuously differentiable for x [, π] and continuously differentiable for t [, T ]. For full credit, you must provide a complete proof of your conclusion. Solution: We first need to shift the boundary condition and let v = u x to obtain v(, t) = v x (π, t) = v(x, ) = φ(x) x. We then use separation of variables v(x, t) = X(x)T (t) to obtain XT = X T X X = T T = λ. Thus we have an equation for X and an equation for T. (a) Find the formal solution

2 i. For the X equation X = λx we have that X(x) = A cos βx + B sin βx (with λ = β ) and the boundary conditions give us X() = A = and These results mean X (π) = = Bβ cos βπ β n = n + 1, n =, 1,.... ii. For the T equation, we have that X n (x) = B n sin((n + 1/)x), n =, 1,.... T = λt which has solution T (t) = Ce λt. iii. Therefore we obtain the formal series solution v(x, t) = B n sin(β n x)e β nt n= iv. By equating this series to the initial data we have that B n = π Thus the formal solution is v(x, ) = φ(x) x = B n sin β n x n= sin(β n x) (φ(x) x) dx. u(x, t) = B n sin(β n x)e β nt + x. (1) n= (b) For the above formal solution to be classical, it is sufficient that φ() =, φ (π) =, φ () =, and φ C 4 [, π]. Convergence of u(x, t): To illustrate the source of these restrictions, first consider the formal series for u(x, t) in Eq. (1). We require this series to be uniformly convergent for (x, t) Ω and the initial/boundary data to be satisfied. For uniform convergence, we invoke the Weierstrass M-test: n a n (x, t) is uniformly convergent if a n (x, t) < M n for (x, t) Ω and n M n <. Since

3 B n sin(β n x)e β nt B n, we require n= B n <. A sufficient condition for this inequality to hold is B n < C n /n for n = 1,,.... We formally compute π π B n = (φ(x) x) sin(β n x)dx = 1 (φ(x) x) cos(β n x) β n = 1 φ() + 1 (φ (x) ) cos(β n x)dx β n β n = 1 φ() + 1 π (φ (x) ) sin(β β n βn n x) 1 β n = 1 φ() + 1 (φ (π) )( 1) n 1 β n βn βn π + 1 (φ (x) ) cos(β n x)dx β n φ (x) sin(β n x)dx φ (x) sin(β n x)dx () If we set φ() = and require φ C [, π], then we can uniformly bound the final integrand φ (x) sin(β n x) < M < and obtain B n < C β n < C n, C = π ( φ (π) + ) + M, n = 1,,..., as needed. Convergence to initial data: Because φ C [, π] and φ() =, φ has a smooth, odd, periodic extension to [ π, π]. Then the Fourier quarter-sine series for φ, u(x, ) x, is guaranteed to converge uniformly to φ(x) for x [, π] by Dirichlet s theorem. Convergence to boundary data: Since each term in the uniformly convergent series (1) evaluates to zero at x =, we have u(, t) =. We check the other boundary condition next. Consider u x. In order to be able to differentiate the series (1) term-by-term, we require each differentiated term to be a) uniformly continuous and b) for the differentiated series S (1) (x, t) = β n B n cos(β n x)e β n t n= to be absolutely summable. If these are both true, then S(x, t) converges uniformly to the continuous function u x (x, t) for (x, t) Ω. The first requirement is satisfied because cos(β n x)e β nt is continuous for all (x, t). Since the domain Ω is compact, each term is uniformly continuous. β n B n < C/n or B n < C/n 3, n = 1,,... is a sufficient condition for absolute summability by the M-test. To obtain sufficient conditions, we resume

4 the calculation in Eq. (), imposing the already specified assumptions π B n = 1 (φ (π) )( 1) n 1 φ (x) sin(β βn βn n x)dx = 1 (φ (π) )( 1) n + 1 π cos(β βn βn 3 n x)φ (x) 1 β n 3 = 1 (φ (π) )( 1) n 1 φ () 1 βn βn 3 βn 3 φ (x) cos(β n x)dx. φ (x) cos(β n x)dx If we set φ (π) = and require φ C 3 [, π], then we can bound the final integrand φ (x) cos(β n x) < M < and obtain (3) B n < C β 3 n < C n 3, C = π φ () + M, n = 1,,.... Therefore S (1) (x, t) converges uniformly to u x (x, t) for (x, t) Ω. We can therefore evaluate the series u x (π, t) = S (1) (π, t) + = as required. Convergence of u xx and u t : The final step is to prove uniform convergence of the series S () (x, t) = βnb n sin(β n x)e β nt. n= By similar arguments as given earlier, B n < C/n 4, n = 1,,... is sufficient for this. We resume the calculation (3) with all prior assumptions π B n = 1 φ () 1 φ (x) cos(β βn 3 βn 3 n x)dx = 1 φ () 1 π φ (x) sin(β βn 3 βn 4 n x) + 1 β n 4 = 1 φ () 1 φ (π)( 1) n + 1 βn 3 βn 4 βn 4 If we set φ () = and require φ C 4 [, π], then φ (x) sin(β n x)dx φ (x) sin(β n x)dx. B n < C β 4 n < C n 4, C = π φ (π) + M, n = 1,,..., and S () (x, t) converges uniformly to continuous u xx (x, t) on Ω. Because u t = u xx, we also have uniform convergence of u t (x, t) to a continuous function on Ω. In summary, if φ() =, φ (π) =, φ () =, and φ C 4 [, π] then u in Eq. (1) is C 1(Ω) and is a classical solution to the initial, boundary value problem.

5 . (Heat equation) Consider the following initial-boundary value problem for the heat equation u t = u xx, x (, 1), t >, u(x, ) = x(1 x), x (, 1), u(, t) = u(1, t) =, t >. Assume the existence of a classical solution u(x, t). (a) Prove the uniqueness of this solution. (b) Show that u(x, t) > on x (, 1) and t >. (c) For each t >, let µ(t) := max x [,1] u(x, t). Show that µ(t) is a nonincreasing function of t. Solution: (a) Assume two solutions u and v, then w = u v solves the problem w t = w xx, x (, 1), t >, w(x, ) =, x (, 1), w(, t) = w(1, t) =, t >. The maximum principle ensures max w(x, t) = and the minimum principle ensures min w(x, t) = for any rectangle = [, 1] [, T ]. Thus w on any, so u v. (b) By the minimum principle, we know min u(x, t) = for any rectangle = [, 1] [, T ], since u(, t) = u(1, t) = < x(1 x) for x (, 1). By the strong minimum principle, we know that if the minimum min u(x, t) = is also obtained at a point (x, t ) for x (, 1) and t (, T ), then u on. However, we know this cannot be since u(x, ) = x(1 x), so u(x, t) must only obtain its minimum on x = and x = 1. (c) By the maximum principle, u(1/, ) = 1/4 = µ(). At each t >, define X(t) such that µ(t) = u(x(t), t). Differentiating, we find µ (t) = u x (X(t), t)x (t) + u t (X(t), t). At (X(t), t), u x (X(t), t) = and u xx (X(t), t). Thus, µ (t) = u xx (X(t), t), so µ(t) is nonincreasing. Alternatively, one could define for each t > an initial boundary value problem with u(x, t ) as the initial condition, so that the maximum principle ensures the maximum on [, 1] [t, T ] lies at t for any T > t >. Thus, u(x, t) u(x, t ) for any t > t >, so µ(t) is nonincreasing.

6 3. (Green s function) Consider the boundary value problem u(x) = f(x), x Ω 3, u(x) = g(x), x Ω. (4) (a) Formulate a boundary value problem for Green s function G(x, y) = Φ(x y) φ x (y) for x Ω using the fundamental solution Φ(x) = (4π x ) 1. (b) Prove that Green s function, if it exists, is unique. (c) Construct Green s function when Ω = B(, 1) { x = (x 1, x, x 3 ) 3 x 3 > }, where B(, 1) is the unit sphere. Solution: (a) Fix x Ω and let G(x, y) = Φ(x y) φ x (y). Then φ x satisfies y φ x (y) =, y Ω, φ x (y) = Φ(x y), y Ω. (b) Fix x Ω and assume the existence of two Green s functions G j (x, y) = 1 4π x y + φ x j (y) for j = 1, where each φ x j satisfies y φ x j (y) =, y Ω, φ x j (y) = Φ(x y), y Ω, j = 1,. Then u(y) = G 1 (x, y) G (x, y) is independent of x and satisfies y u(y) =, y Ω, u(y) =, y Ω, i.e., u is harmonic in Ω. By the maximum/minimum principles and the boundary data, u(y) so that G 1 (x, y) = G (x, y) and Green s function is unique. (c) Fix x Ω and let G(x, y) = Φ(x y) φ x (y). First, we construct Green s function for the unit sphere B(, 1) with the method of images. Let φ x B(y) = Φ( x ( x y)), x = x x. Since x is not in B(, 1), φ x B(y) is harmonic in B(, 1). Let y = 1, then we compute x ( x y) x = x x y = 1 x y + x = x y.

7 Then, for y B(, 1), we have φ x B(y) = Φ( x ( x y)) = Φ(x y), as required and Green s function for the unit sphere is G B (x, y) = Φ(x y) φ x B(y). We now use the method of images again to obtain Green s function for the upper hemisphere Ω by reflecting about the plane x 3 =. Let G(x, y) = G B (x, y) G B (ˆx, y), ˆx = (x 1, x, x 3 ) = Φ(x y) Φ( x ( x y)) Φ(ˆx y) + Φ( ˆx ( ˆx y)) = Φ(x y) φ x (y). Since x, ˆx, and ˆx are not in Ω, φ x (y) is harmonic for y Ω. We need to show that φ x (y) = Φ(x y) for y Ω. If y = 1 then as before, x x y = x y. Similarly, ˆx ˆx y = ˆx y. Then, for y = 1 and y 3, φ x (y) = Φ( x ( x y)) + Φ(ˆx y) Φ( ˆx ( ˆx y)) = Φ(x y) + Φ(ˆx y) Φ(ˆx y) = Φ(x y), as required. Finally, if y = (y 1, y, ), y 1 + y 1, then x x y = ˆx ˆx y and ˆx y = x y so that φ x (y) = Φ( x ( x y)) + Φ(ˆx y) Φ( ˆx ( ˆx y)) = Φ( ˆx ( ˆx y)) + Φ(x y) Φ(ˆx y) = Φ(x y), and the construction is complete.

8 4. (Wave equation) Consider the wave equation u tt = c u xx, x, t >, u(x, ) = φ(x), x, u t (x, ) = ψ(x), x. (5) Assume the existence of a classical solution u(x, t). (a) If φ(x) and ψ(x) are both odd functions of x, show that the solution u(x, t) is odd in x for t >. (b) Find a solution to Eq. (5) assuming φ(x) and ψ(x), and prove that it is unique. (c) Assume φ(x) and ψ(x) have compact support, fix c = 1, and define the kinetic K(t) = 1 u t(x, t) dx and potential P (t) = 1 u x(x, t) dx energies. Show that K(t) = P (t) for all t sufficiently large. Solution: (a) By d Alembert s formula u( x, t) = 1 [φ( x + ct) + φ( x ct)] + 1 c = 1 [ φ(x ct) φ(x + ct)] 1 c ( 1 = [φ(x ct) + φ(x + ct)] + 1 c x+ct x ct x ct x+ct x+ct x ct ψ(s)ds ψ( s)ds ) ψ(s)ds = u(x, t) (b) u is a solution since u(x, ) φ(x); u t (x, ) ψ(x); u tt = = u xx. Define the energy E(t) = 1 u t(x, t) + c u x (x, t) dx = Ē, constant due to conservation. Thus, since E() = 1 φ(x) + c ψ(x) dx = = Ē. Since u t, u x, this implies u t u x, and since u(x, ), then u(x, t). (c) By d Alembert s formula so u(x, t) = 1 [φ(x + t) + φ(x t)] + 1 x+t x t ψ(s)ds u t (x, t) = 1 [φ (x + t) φ (x t)] + 1 [ψ(x + t) + ψ(x t)] and u x (x, t) = 1 [φ (x + t) + φ (x t)] + 1 [ψ(x + t) ψ(x t)].

9 Now, since both φ and ψ have compact support, for sufficiently large t, all terms = φ (x + t)φ (x t)dx φ (x + t)ψ(x t)dx φ (x t)ψ(x + t)dx ψ(x t)ψ(x + t)dx. That is, if φ(x) is supported on x [a, b] and ψ(x) is supported on x [m, n], then once t > max(n m, b a, n a, b m)/ = t, the length t will be wider than the region containing the support of both functions, so φ (x) and ψ(x) must be zero either at x + t or x t. Thus, we compute K(t) = 1 8 [ φ (x + t) + φ (x t) + φ (x + t)ψ(x + t) φ (x t)ψ(x t) + ψ(x + t) + ψ(x t) ]dx. For any finite t > t, we can change variables of all terms z = x+t and z = x t and not change the integral, so Similarly, P (t) = 1 8 and changing variables, so K(t) = P (t) for t > t. K(t) = 1 4 [ φ (z) + ψ(z) ] dz. [ φ (x + t) + φ (x t) + φ (x + t)ψ(x + t) φ (x t)ψ(x t) + ψ(x + t) + ψ(x t) ]dx. P (t) = 1 4 [ φ (z) + ψ(z) ] dz,

10 5. (Method of characteristics) Consider the quasilinear equation (y + u)u x + yu y = x y. (a) Give an example of a connected curve Γ such that the Cauchy problem with prescribed data on that curve cannot be solved. (b) Given the Cauchy data u(x, 1) = 1 + x. What are the characteristic curves? Find the solution. For what values of (x, y) does the solution exist? Solution: (a) We require the data u(x (s), y (s)) = u (s), s I to lie on a characteristic, i.e., for the vector field v = (y +u, y) to be tangent to the initial curve. For example, v is tangent to the curve Γ = {(x, ) x } for any choice of initial data. Note that the zero Jacobian condition x J = (s) y (s) y (s) + u (s) y (s) = x y y (y + u ) = for all s I, is necessary but not sufficient to prove that the problem cannot be solved. (b) We parametrize the data as x = s, y = 1, u = 1 + s for s. characteristic equations for x(ξ, s), y(ξ, s), and u(ξ, s) are x ξ = y + u, x(, s) = s, y ξ = y, y(, s) = 1, u ξ = x y, u(, s) = 1 + s, s. Taking another derivative of the first equation, we obtain x ξξ = y ξ + u ξ = y + x y = x, Then the which has solution x = A(s)e ξ + B(s)e ξ. The initial data imply x(, s) = A(s) + B(s) = s and x ξ (, s) = A(s) B(s) = y(, s) + u(, s) = + s. Then A(s) = 1 + s and B(s) = 1 so that The equation for y is solved Then the characteristic curves are x(ξ, s) = (1 + s)e ξ e ξ. y(ξ, s) = e ξ. x = (1 + s)y 1 y

11 for each s. We solve the equation for u u ξ = se ξ e ξ u(ξ, s) = se ξ + e ξ. Undoing the characteristic transformation, we obtain the solution which exists for y. u(x, y) = x y +, x, y, y